Chapter 5 : Analysis Methods

Example 5.8 Page No : 76

In [1]:
import math 
#From figure 5.13(a)

#Applying KVL equation to the loop 
I = (20.+10)/(3+6)

# Calculation and Results
#As current will not flow in upper 3 ohm resistor so Thevenin voltage is equal to either of the two parallel branches
V1 = 20.-I*3
print "Thevenin voltage  =  %dV"%(V1)

# Left 3 ohm and 6 ohm resistor are in parallel and their equivalent is in series with 3 ohm
R1 = 3+(3.*6)/(3+6)
print "Thevenin resistance  = %dohm"%(R1)

#Now to find Norton's equivalent
I1 = V1/R1
print " Norton current  = %dA"%(I1)
#The value of resistance in Norton equivalent will not change but will come in parallel with current source")
Thevenin voltage  =  10V
Thevenin resistance  = 5ohm
 Norton current  = 2A